well, we know the angle θ terminal location, is at -3,2, so the adjacent side is -3 and the opposite side is 2, so it looks like the picture below, therefore
[tex]\bf (\stackrel{a}{-3}~,~\stackrel{b}{2})\impliedby \textit{let's find the \underline{hypotenuse}}
\\\\\\
\textit{using the pythagorean theorem}
\\\\
c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}
\qquad
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
c=\sqrt{(-3)^2+2^2}\implies c=\sqrt{13}\qquad \qquad \qquad
sec(\theta )=\cfrac{\stackrel{hypotenuse}{\sqrt{13}}}{\stackrel{adjacent}{-3}}[/tex]
